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3 years ago

How many grams are in 6.8L of Oxygen gas (O2) at STP

Chemistry

1 answer:

zubka84 [21]3 years ago

4 0

Answer:

9.7 g

Explanation:

From the question,

Note: The molar volume of all gas at stp is 22.4 dm³ or 22.4 L

1 mol of oxygen gas (O₂) at stp = 22.4 dm³

X mole of oxygen gas (O₂) at stp = 6.8 L

X = (1 mol×6.8 L)/22.4 L

X = 0.3036 mol.

But,

Number of mole (n) = mass (m)/molar mass (m')

n = m/m'

m = n×m'.................. Equation 2

Where n = 0.3036 mol, m' = 32 g/mol

Substitute into equation 2

m = 0.3036×32

m = 9.7 g

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Answer:

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Explanation:

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5 0

3 years ago

How many grams of CO are needed to react with an excess of Fe 2 O 3 to produce 209.7 g Fe?

gavmur [86]

Answer:

Mass = 157.5 g

Explanation:

Given data:

Mass of CO needed = ?

Mass of Fe formed = 209.7 g

Solution:

Chemical equation:

3CO + F₂O₃ → 2Fe + 3CO₂

Number of moles of Fe:

Number of moles = mass/ molar mass

Number of moles = 209.7 g/ 55.85 g/mol

Number of moles = 3.75 mol

Now we will compare the moles of iron and carbon monoxide.

Fe : CO

2 : 3

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Mass = number of moles × molar mass

Mass = 5.625 mol × 28 g/mol

Mass = 157.5 g

4 0

3 years ago

A sample of N2 gas in a flask is heated from 27 Celcius to 150 Celcius. If the original gas is @ pressure of 1520 torr, what is

Romashka [77]

Answer:

$\large \boxed{\text{B.) 2.8 atm}}$

Explanation:

The volume and amount are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

$\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}$

Data:

p₁ = 1520 Torr; T₁ = 27 °C

p₂ = ?; T₂ = 150 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = ( 27 + 273.15) K = 300.15 K

T₂ = (150 + 273.15) K = 423.15 K

(b) Calculate the new pressure

$\begin{array}{rcl}\dfrac{1520}{300.15} & = & \dfrac{p_{2}}{423.15}\\\\5.064 & = & \dfrac{p_{2}}{423.15}\\\\5.064\times423.15&=&p_{2}\\p_{2} & = & \text{2143 Torr}\end{array}\\$

(c) Convert the pressure to atmospheres

$p = \text{2143 Torr} \times \dfrac{\text{1 atm}}{\text{760 Torr}} = \textbf{2.8 atm}\\\\\text{The new pressure reading will be $\large \boxed{\textbf{2.8 atm}}$}$